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For the given matrix A, one of the Eigenvalue is real

$A=\begin{bmatrix} 1 &2 &3 &4 &5 \\ 5 &1 &2 &3 &4 \\ 4&5 &1 &2 &3 \\ 3&4 &5 &1 &2 \\ 2 &3 &4 &5 &1 \end{bmatrix}$

The real Eigen value is:
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Ans -1

$\begin{bmatrix} 1 & 2 &3 & 4 & 5\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

=$\begin{bmatrix} 15 & 15 &15 & 15 & 15\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

[where $R_{1}\leftarrow R_{1}+R_{2}+R_{3}+R_{4}+R_{5}$]

=$15\begin{bmatrix} 1 & 1 &1 & 1 & 1\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

=$15\begin{bmatrix} 1 & 1 &1 & 1 & 1\\ 14& 13 & 12 & 11 & 10\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

[Here $R_{2}\leftarrow R_{2}+R_{3}+R_{4}+R_{5}$]

=$15\begin{bmatrix} 0 & 0 & 0 & 0 &1 \\ 1 & 1 &1 &1 & 10\\ -1&4 & -1 &-1 & 3\\ -1 & -1 & 4 &-1 &2 \\ -1 & -1 &-1 &4 & 1 \end{bmatrix}$

[Here $C_{1}\leftarrow C_{1}-C_{2}$

$C_{2}\leftarrow C_{2}-C_{3}$

$C_{3}\leftarrow C_{3}-C_{4}$

$C_{4}\leftarrow C_{4}-C_{5}$]

=$15\times 1\begin{bmatrix} 1 & 1 & 1 &1 \\ -1& 4 & -1 &-1 \\ -1 &-1 &4 &-1 \\ -1& -1 &-1 &4 \end{bmatrix}$

=$15\times 1\begin{bmatrix} 0& 0 & 0 &1 \\ -5& 5 &0 &-1 \\ 0 &-5 &5 &-1 \\ 0& 0 &-5 &4 \end{bmatrix}$

=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1& 1 &0 \\ 0 &-1 &1 \\ 0& 0 &-1 \end{bmatrix}$

=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1-\lambda & 1 &0 \\ 0 &-1-\lambda &1 \\ 0& 0 &-1-\lambda \end{bmatrix}$

=(-1-$\lambda$)(-1-$\lambda$)^2

$\lambda =-1$

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How eleminating column

$\begin{bmatrix} 0 &0 & 1\\ x& y &z \\ p& q &r \end{bmatrix}$

=$0\begin{bmatrix} y & z\\ q & r \end{bmatrix}$-$0\begin{bmatrix} x & z\\ p& r \end{bmatrix}$+$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$

=$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$

So, 1 row, and 1 column eleminated
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Every row sum equals $15$, so $15$ is an eigenvalue.
Answer:

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